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Ch 7.3 Using Chemical Formulas

Ch 7.3 Using Chemical Formulas. The Mass of a Mole of an Element. Remember: The atomic mass of an element (a single atom) is expressed in atomic mass units (amu). Molar Mass: is the atomic mass of an element expressed in grams/mole (g/mol). Carbon = 12.01 g/mol Hydrogen = 1.01 g/mol

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Ch 7.3 Using Chemical Formulas

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  1. Ch 7.3 Using Chemical Formulas

  2. The Mass of a Mole of an Element • Remember: The atomic mass of an element (a single atom) is expressed in atomic mass units (amu). • Molar Mass: is the atomic mass of an element expressed in grams/mole (g/mol). • Carbon = 12.01 g/mol • Hydrogen = 1.01 g/mol • When dealing with molar mass, round off to two decimals. 12.011g/mol -> 12.01g/mol

  3. The Mass of a Mole of a Compound • You calculate the mass of a molecule by adding up the molar masses of the atoms making up the molecules. • Example: H2O • H = 1.01 g x 2 atoms = 2.02 g/mol • O = 16.00 g x 1 atom = 16.00 g/mol • Molar Mass of H2O = 2.02 g/mol + 16.00 g/mol = 18.02 g/mol • This applies to both molecular and ionic compounds

  4. Find the molar mass of PCl3 • P = 30.97 g x 1 atom = 30.97 g/mol • Cl = 35.45 g x 3 atoms = 106.35 g/mol • PCl3 = 30.97 g + 106.35 g = 137.32 g/mol • What is the molar mass of Sodium Hydrogen Carbonate (NaHCO3) ? • Na = 22.99 g x 1 atom = 22.99 g/mol • H = 1.01 g x 1 atom = 1.01 g/mol • C = 12.01 g x 1 atom = 12.01 g/mol • O = 16.00 g x 3 atoms = 48.00 g/mol • NaHCO3 = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 g/mol

  5. Converting Moles to Mass • You can use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance. • Mass (g) = # of moles x mass (g) 1 mole Example: If molar mass of NaCl is 58.44 g/mol, what is the mass of 3.00 mol NaCl? Mass of NaCl = 3.00 mol x 58.44g = 1 mol 175 g NaCl

  6. Example 2: Moles to Mass • What is the mass of 9.45 mol of Aluminum Oxide (Al2O3)? • Find molar mass of Al2O3 = 101.96 g/mol • Mass = 9.45 mol Al2O3 x 101.96 g Al2O3 1 mol Al2O3 = 964 g Al2O3

  7. Converting Mass to Moles • You can invert the conversion factor to find moles when given the mass. • Moles = mass (g) x 1 mole mass (g) Example: If molar mass of Na2SO4 142.05 g/mol, how many moles is 10.0 g of Na2SO4? Moles of Na2SO4 = 10.0 g x 1 mol = 142.05 g = 0.0704 mol Na2SO4

  8. Example 2: Mass to Moles • How many moles are in 75.0 g of Dinitrogen Trioxide? • Find molar mass of N2O3 = 76.02 g/mol • Moles = 75.0 g N2O3 x 1 mole = 76.02 g N2O3 0.987 mol N2O3

  9. Percent Composition • Percent Composition: the relative amount of the elements in a compound. • Also known as the percent by mass • It can be calculated in two ways: • Using Mass Data • Using the Chemical Formula % mass of element= mass of element x100% mass of compound

  10. Example • When a 13.60 g sample of a compound containing Mg and O is decomposed, 5.40 g O is obtained. What is the % composition of this compound? Mass of compound: 13.60 g Mass of oxygen: 5.40 g O Mass of magnesium: 13.60 g - 5.40 g = 8.20 g Mg % Mg = 8.20 g Mg x 100% = 13.60 g % O = 5.40 g O x 100% = 13.60 g 60.3% 39.7%

  11. Find the percent composition of Cu2S. • Find mass of Cu and S • Cu = 63.55 x 2 = 127.10 g • S = 32.07 g • Find mass of Cu2S • 127.10 g + 32.07 g = 159.17 g % Composition • Cu = 127.10 g x 100% = 159.17 g • S = 32.07 g x 100% = 159.17 g 79.85% 20.15%

  12. Homework • 7.3 pg 253 #30-33

  13. Ch 7.4Determining Chemical Formulas

  14. Empirical Formulas • Empirical Formula: shows the smallest whole-number ratio of the atoms of the elements in a compound. • Example: • The Empirical Formula for Hydrogen Peroxide (H2O2) is HO with a 1:1 ratio. • The Empirical Formula for Carbon Dioxide (CO2) is CO2 with a 1:2 ratio.

  15. Determining the Empirical Formula of a Compound • A compound is found to contain 25.9% Nitrogen and 74.1% Oxygen. What is the Empirical Formula of the compound? • 25.9 g N x 1 mol N = 14.01 g N • 74.1 g O x 1 mol O = 16.00 g O • N1.85O4.63 = N1O2.5 = N2O5 1.85 mol N 4.63 mol O

  16. Molecular Formulas • Molecular Formula: tells the actual number of each kind of atom present in a molecule of a compound • Example: • The Molecular Formula for Hydrogen Peroxide is H2O2. • The Molecular Formula for Carbon Dioxide is CO2 • It is possible to find the Molecular Formula using the Empirical Formula if you know the molar mass of the compound.

  17. Finding the Molecular Formula • Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N • Step 1: Find the empirical formula molar mass • 12.01 + (4 x 1.01) + 14.01 = 30.06 g/mol • Step 2: Divide molar mass by EF molar mass • 60.0 g/mol = 1.99  2 30.06 g/mol • Step 3: Multiply empirical formula by 2 • CH4N x 2 = C2H8N2

  18. Homework • 7.4 pg 253 #36-38

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